Essentials of Genetics (9th Edition) - Standalone book
9th Edition
ISBN: 9780134047799
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino
Publisher: PEARSON
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Chapter 3, Problem 3PDQ
When working out genetics problems in this and succeeding chapters, alwaysassumethatmembersof the P1 generation are homozygous, unless the information or data you are given require you to do otherwise.
In a cross between a black and a white guinea pig, all members of the F1 generation are black. The F2 generation is made up of approximately 3/4 black and 1/4 white guinea pigs. Diagram this cross, and show the genotypes and
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When working out genetics problems in this and succeeding chapters, always assume that members of the P1 generation are homozygous, unless the information or data you are given require you to do otherwise.HOW DO WE KNOW? we focused on the Mendelian postulates, probability, and pedigree analysis. We also considered some of the methods and reasoning by which these ideas, concepts, and techniques were developed. On the basis of these discussions,
what answers would you propose to the followingquestion.
Question: Since experimental crosses are not performed in humans, how do we know how traits are inherited?
Required information
A single-factor cross is one in which the inheritance of only one character and its associated genotypes are followed.
Punnett squares are often used to predict the outcomes of simple genetic crosses. Based on Mendel's laws, the
genotypes of the parents can be used to predict the genes in their gametes and the resulting progeny. A Punnett square
enables you to predict the types of offspring the parents are expected to produce and in what proportions.
Sickle cell anemia is a recessive trait in humans. In a cross between two parents who are heterozygous for the gene, what are the gamete possibilities of the parer
Mother's gamete
possibilities
Father's gamete
possibilities
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Question: On the basis of Mendel's hypothesis and observations, predict the results from the following crosses in garden peas: (a) a tall (dominant and homozygous) variety crossed with a dwarf variety: (b) the progeny of (a) selfed; (c) the progeny from (a) crossed with the original tall parent; (d) the progeny from (a) crossed with the original dwarf-parent variety.
Chapter 3 Solutions
Essentials of Genetics (9th Edition) - Standalone book
Ch. 3 -
CASE STUDY | To test or not to test
Thomas...Ch. 3 -
CASE STUDY | To test or not to test
Thomas...Ch. 3 - CASE STUDY | To test or not to test Thomas first...Ch. 3 -
CASE STUDY | To test or not to test
Thomas...Ch. 3 - When working out genetics problems in this and...Ch. 3 - When working out genetics problems in this and...Ch. 3 - When working out genetics problems in this and...Ch. 3 -
When working out genetics problems in this and...Ch. 3 - When working out genetics problems in this and...Ch. 3 -
When working out genetics problems in this and...
Ch. 3 - When working out genetics problems in this and...Ch. 3 - When working out genetics problems in this and...Ch. 3 - When working out genetics problems in this and...Ch. 3 - When working out genetics problems in this and...Ch. 3 -
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When working out genetics problems in this and...Ch. 3 - When working out genetics problems in this and...Ch. 3 - When working out genetics problems in this and...Ch. 3 -
When working out genetics problems in this and...Ch. 3 - When working out genetics problems in this and...Ch. 3 -
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- Question: This is a normal 3 point test cross, except that instead of regular phenotypes, you are looking at DNA markers on a gel. One parent, according to the gel, is heterozygous at each marker. The other parent is homozygous for each marker. (Again, this means it is a test cross: AaEeHh x AAEEHH --but don't be confused by that, because these are not "dominant" and "recessive" per se; the phenotype is just a band on a gel). For each offspring, figure out its genotype (homozygous or heterozygous for each gene. Then, figure that one parent made only AEH gametes, so you can cross that out if it helps.) Then treat it pretty much the same as a 3 point test cross.arrow_forwardNow cross two of the F₁ offspring. Parent 1 Gametes F2 Offspring Parent 2 Gametes 24. What is the phenotypic ratio in the F2 generation? 25. In the dihybrid cross you have considered two traits at a time. Although the number of traits has increased by one, what has happened to the number of possible phenotypes of offspring produced in the F2 generation? ni llit asa2013 bhidydenom sih at Simons 1:11 sabrax gatamearrow_forwardGenetic Problems Read the following genetic problems, and then complete the activities and questions for each problem 1. In the pea plant, the green pod color allele is dominant to the yellow pod color allele. o Set-up a monohybrid cross between a pea plant that is heterozygous and a true-breeding yellow pod pea plant. o What percentage of the offspring produced from this cross would you expect to have yellow pods? 2. Within a mouse population, the black fur allele (B) is dominant to the white fur allele (b) and the short whisker allele (S) is dominant to the ngs long-whisker allele (s). o A heterozygous black-furred short-whiskered mouse is crossed with a homozygous white-furred long-whiskered mouse. o What percentage of the offspring will be black-furred with long whiskers? 3. A physician is examining the blood types of children from a family. o Child 1 has blood type AB, Child 2 has blood type B, Child 3 has blood type B, and Child 4 has blood type A. o Based on the phenotypes of…arrow_forward
- Instruction - Please answer them correctly - Please answer all of them, they are connected. PEDIGREE ANALYSIS and SYMBOLOGY Examine the pedigree which has X linked Dominant inheritance of disorder. Use letter X* (asterisk denotes disorder) as genotype of the individuals which can be XX, XY, X*X*, X*X and X*Y. a. What is the genotype of IV-6? b. What is the genotype of III-6? c. What is the genotype of II-3? d. What is the genotype of III-8? e. If couple I-1 and I-2 will have a son, what is the probability of having the disorder? f. If couple III-8 and III-9 will have another child, what is the probability of having the disorder? g. Theoretically, if individual IV-3 and individual IV-5 will marry and will have a child, what is the probability of having a child without the X-linked disorder?arrow_forwardMonohybrid Problems llustrating Codominamce 5. Another type of monohybrid inheritance involves the expression of both phenotypes in the heterozy- gous situation. This is called codominance. One well-known example of codominance occurs in the coat color of Shorthorn cattle. Those with reddish-gray roan coats are heterozygous (RR'), and result from a mating between a red (RR) Shorthorn and one that's white (R'R'). Roan cattle don't have roan-colored hairs, as would be expected with incom- plete dominance, but rather appear roan as a result of having both red and white hairs. Thus, the roan col- oration is not a consequence of pigments blending in each hair. Because the R and R' alleles are both fully expressed in the heterozygote, they are codominant. a. If a roan Shorthorn cow is mated with a white bull, what will be the genotypic and phenotypic ratios in the F, generation? genotypic ratio . phenotypic ratio b. List the parental genotypes of crosses that could produce at least some…arrow_forward11:41 Cancel Markup Done Name: Date: Monohybrid practice problems In pea plants, the traits below exhibit the following dominance patterns: Recessive Expression: Wrinkled Dominant Expression: Round Purple |Yellow Inflated Green Trait: 1. Seed shape (R) 2. Flower color (P) 3. White Green Constricted Yellow Terminal Short Color of seed coat (Y) Form of ripe pods (I) 4. 5. |Color of unripe pods (G) 6. Position of flowers (A) 7. Length of stem (T) Axial Tall Record the genotypes for pea plants with the following descriptions (The first one has been done for you: 1. а. дg A plant with yellow pods A planteozygous for ereen pods С. A plant homozvaoue ta vellow seeds A plant with white flowers A plant with areen seeds 2. Complete the Punnett Square showing the cross between a pea plant with pure round seeds and a plant with wrinkled seeds. Summarize the phenotypes and genotypes for the offspring. Parental cross Genotypic Percentages: Phenotypic Percentages: 3. A pea plant with pure yellow…arrow_forward
- 11:42 Cancel Markup Done Name: Date: Monohybrid practice problems In pea plants, the traits below exhibit the following dominance patterns: Recessive Expression: Wrinkled Dominant Expression: Round Purple |Yellow Inflated Green Trait: 1. Seed shape (R) 2. Flower color (P) 3. White Green |Constricted Yellow Terminal Short Color of seed coat (Y) Form of ripe pods (I) 4. 5. |Color of unripe pods (G) 6. Position of flowers (A) 7. Length of stem (T) Axial Tall Record the genetvnes f-Der-- = foltrawino deserintins (The first one has been done for you): а. дg A plant with yellow poás- EplancteOZygous for ereen pods N 0)e Owers С. A plant homozvaoue ycilow seeds A plant with white flowers TA piamt wilh areen seeds. Complete the Punnett Square showing the eress between a pea plant with pure round seeds and a plant with wrinkled seeds. Summanze the phenotypes and genotypes for the offspring. Genolynic PorcentadCS. Parental cross PhenolypicPerceniageS 3A pea plant with pure velow seeds is erossed…arrow_forwardQuestion GROUP A: Genetics Problems In humans, red-green color blindness is caused by a recessive allele on the X chromosome. A male with normal vision and his color-blind wife have a child. If the child is male what is the probability that he will be color-blind? The female in the previous problem has a daughter that is color-blind. The husband claims that this cannot be his child. Can you support or refute his argument?arrow_forward12:52 0: 9 59.0 a 4l 4 96% KB/S bartleby.com/questions-and-answ = bartleby Science / Biology / Q&A Library / Imagine you h Imagine you have a blood group c Start your trial now! First week only $4.99! → Question Imagine you have a blood group of "X" which is recessive and expressed by xx. The dominant blood groups are Y and Z, where homozygous of these alleles are expressed as YY and ZZ, respectively. What will be the genotype of your parents blood group? Why? Please explain in your own words. [Max 200 words)] View transcribed image text ) Expand Expert Answer Want to see the step-by-step answer?arrow_forward
- help needed with all of them please A pesticide resistance allele was at a frequency of 50% in a population of flies. The pesticide was no longer effective because of the prevalence of resistant flies. Ranchers decided to stop using the pesticide, and in 20 years the resistance allele dropped to a frequency of 5%. The allele is autosomal, affects fitness in an additive fashion, and there are two fly generations per year. a) Explain in 1—2 sentences why natural selection may have caused a decrease in the frequency of the resistance allele. b) What is the selection coefficient acting against homozygotes for the resistance allele (in the absence of the pesticide)? c) What was the frequency of the resistance allele one generation after ranchers stopped using the pesticide (i.e., after one generation of viability selection)? Assume that the population was in Hardy-Weinberg equilibrium when ranchers stopped using the pesticide and the population underwent random mating. d) The effective…arrow_forwardQuestion- 1. In onions, male sterility is due to the interaction of a chromosomal allele pair hh and “sterile” (S) cytoplasm. All other combinations (i.e. HH/Hh and “sterile” cytoplasm, HH/Hh or hh “fertile” (F) cytoplasm) result in male-fertile plants. The male-sterile trait is incorporated into inbred lines to produce hybrid F1 seed on a commercial scale.a) How would you perpetuate the male-sterile line? Show the cross.b) Briefly outline the method of producing hybrid seed (heterozygote) for the commercial crop. Show the cross.c) Does it matter whether the cytoplasm is fertile or sterile in the male-fertile inbred? Explain.arrow_forwardQuestion: Black hair of the mouse is produced by dominant gene B and white hair by recessive allele b.Female with black hair crossed with male black hair (both genotypes are heterozygous). Produced individuals 112 and 113, female with white hair and male with black hair, then mating occur between 114 and 113, and another mating occur between 112 and 111. assume that 111 and 114 individuals do not carry the recessive allele, represents the above in a pedigree chart and determine the genotype of each individual.arrow_forward
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